Question

Compute 2j < (3i - 4k) + (i + 2j) k.
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Answer 1

Answer:

i× i = 0, j ×j = 0, k × k = 0.

j×i = (-k), k×j = (-i), i×k = (-j)

2j × (3i - 4k) + (i × 2j) × k.

= -6k -8i + (-j) + 2i.

= -6i - j - 6k.

= -(6i + j + 6k) .